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CIE 9231 2023 June Paper 13 Q1

A Level / CIE / FP1

CIE 9231 2023 June Paper 13 Paper · Question 1

题目

Problem

Prove by mathematical induction that, for all positive integers nn, 53n+32n−335^{3n}+32^n-33 is divisible by 3131.

(6)
题目中文翻译

用数学归纳法证明,对于所有正整数 nn,53n+32n−335^{3n}+32^n-33 都能被 3131 整除。

(6)
原卷通用作答要求

Answer all questions. You should use a calculator where appropriate.

You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.

Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.

请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。

解答

解法一

思路

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先验n=1,再比较相邻表达式,差的两个系数都是31的倍数。

答题过程

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Let Fn=53n+32n−33F_n=5^{3n}+32^n-33. For n=1n=1,

F1=125+32−33=124=31×4,F_1=125+32-33=124=31\times4,

so the statement holds.

Assume that FkF_k is divisible by 3131 for some positive integer kk. Then

Fk+1−Fk= (53−1)53k+(32−1)32k= 124⋅53k+31⋅32k= 31(4⋅53k+32k).\begin{align*} F_{k+1}-F_k =&\,(5^3-1)5^{3k}+(32-1)32^k\\[4mm] =&\,124\cdot5^{3k}+31\cdot32^k\\[4mm] =&\,31\big(4\cdot5^{3k}+32^k\big). \end{align*}

This difference is divisible by 3131. Since FkF_k is divisible by 3131, so is Fk+1F_{k+1}.

The statement is true for n=1n=1 and truth for n=kn=k implies truth for n=k+1n=k+1. Thus, by mathematical induction, 53n+32n−335^{3n}+32^n-33 is divisible by 3131 for every positive integer nn.